Questions on sound were discussed earlier on this site. You can access all posts related sound by clicking on the label ‘sound’ below this post.

Today we will discuss a few more multiple choice practice questions on sound, relevant to AP Physics B aspirants.

(1) A source producing sound of wave length 0.6 m is moving away from a stationary listener with speed

*V*/6 where*V*is the speed of sound in air. The wave length of sound heard by the listener is
(a) 0.5 m

(b) 0.54 m

(c) 0.66 m

(d) 0.7m

(e) 0.8 m

You might have come across questions of the above type under Doppler effect.

If the real frequency of the source of sound is

*n*we have*V/n =*

*λ*where

*λ*is the wavelength of the sound as measured by the stationary listener when the source is stationary. Therefore we have

*V/n =*0.6 ……………..(i)

When the source moves away from the listener at speed

*V*/6 the*n*sound waves produced per second will occupy a distance*V + V/*6 so that the wave length as measured by the listener becomes (*V + V/*6)/*n*. Therefore we have
(

*V + V/*6)/*n*=*λ*_{1}……….(ii)
where

*λ*_{1}is the new wave length.
Dividing Eq (ii) by Eq (i) we obtain

1 + 1/6 =

*λ*_{1}/0.6
Or, 7/6 =

*λ*_{1}/0.6 from which*λ*_{1}= 0.7 m.
(2) Two steel wires A and B of the same length are vibrating under the same tension. If the first overtone of wire A has the same frequency as the fundamental note of wire B, the area of cross section of wire A is

(a) twice that of B

(b) three times that of B

(c) four times that of B

(d) half that of B

(e) one third that of B

The frequency of vibration (

*f*)*of a string of length**ℓ*and linear density*m*stretched under tension*T*is given by*f*= (

*s*/2

*ℓ*) √(

*T/m*) where

*s*= 1,2,3,4 etc. For the fundamental mode

*s*= 1. For the 2

^{nd}mode (or 1

^{st}overtone)

*s*= 2. For the 3

^{rd}mode (or 2

^{nd}overtone)

*s*= 3 and so on.

The linear density is mass per unit length and is equal to

*a**ρ*where ‘*a*’ is the area of cross section and*ρ*is the density of the material of the string.
Since the first overtone of wire A has the same frequency as the fundamental note of wire B, we have

(2/2

*ℓ*) √(*T/m*_{1}) = (1/2*ℓ*) √(*T/m*_{2}) where*m*_{1}is the linear density of A and*m*_{2}is the linear density of B.
From mthe above equation it follows that

*m*_{1 }= 4*m*_{2}.
Since the wires are of the same material, the area of cross section of A must be 4 times that of B.

[Since the changes are confined to the mode of vibration and the linear density, you should be able to write 2/√

*m*_{1}= 1/√*m*_{2}to arrive at*m*_{1 }= 4*m*_{2}and the final answer*a*_{1}= 4*a*_{2}]
(3) Two sound notes of wave lengths

*λ*_{1}and*λ*_{2}in air produce*n*beats per second. If*λ*_{1}<*λ*_{2}the speed of sound in air is
(a)

*λ*_{1}*λ*_{2}*n /*(*λ*_{2}+*λ*_{1})
(b) (

*λ*_{1 }+*λ*_{2})*n*
(c) (

*λ*_{1}*λ*_{2}*/n*)(*λ*_{2}+*λ*_{1})
(d)

*λ*_{1}*λ*_{2}*n /**λ*_{1}
(e)

*λ*_{1}*λ*_{2}*n /*(*λ*_{2}–*λ*_{1})
The frequency

*f*_{1}*of the sound of wave length**λ*_{1}is given by*f*

_{1 }=

*v/λ*

_{1}where

*v*is the speed of sound in air.

The frequency

*f*_{2 }of the sound of wave length*λ*_{2}is given by*f*

_{2 }=

*v/λ*

_{2}

Since

*λ*_{1}<*λ*_{2 }we have*f*_{1}>*f*_{2}
Since the number of beats produced per second is

*n*, we have*f*

_{1}–

*f*

_{2}=

*n*

Therefore,

*v/λ*_{1}–*v/λ*_{2}=*n*
Or,

*v*(1*/λ*_{1}– 1*/λ*_{2}) =*n*
This gives

*v*=*λ*_{1}*λ*_{2}*n /*(*λ*_{2}–*λ*_{1})
(4) A glass tube is open at both ends. The minimum frequency of a tuning fork which vibrates in resonance with the air column in this tube is

*f*. If this tube is held vertically with half its length submerged in water, what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube?
(a)

*f*/3
(b)

*f/*2
(c)

*f*
(d) 2

*f*
(e) 3

*f*
The air column will vibrate with the

*minimum*frequency in the fundamental mode. When both ends of the glass tube are open, there are anti-nodes at the ends and the adjacent node in the middle so that the length*L*of the tube is equal to*λ/*2 where*λ*is the wave length of the fundamental note. Thus we have*L = λ/*2

Or

*λ =*2*L*
[Remember that the distance between consecutive anti-nodes (or, consecutive nodes) is equal to

*λ/*2].
When half the length of the glass tube is immersed in water, it becomes a closed pipe of length

*L/*2. In the fundamental mode of vibration of the air column in the tube in this case there is a node at the closed end (water surface) and the adjacent anti-node at the open end as shown in the figure. Therefore we have*L/*2

*= λ*

_{1}

*/*4 where

*λ*

_{1}is the wave length of the resonating sound.

Or

*λ*_{1}*=*2*L*
In both cases the wave length of the resonating sound is the same. Therefore the resonant frequency is the same as

*f*[Option (c)].
(5) In the above question what will be the minimum frequency of a tuning fork that will vibrate in resonance with the air column in the tube if a quarter of the tube is submerged in water?

(a) 2

*f*/3
(b) 3

*f/*4
(c)

*f*
(d) 2

*f*
(e) 3

*f*
Initially, when both ends are open, the resonating wavelength as shown above is given by

*λ =*2

*L*

When a quarter of the glass tube is submerged in water, we have

3

*L/*4*= λ*_{1}/4
Therefore

*λ*_{1}*=*3*L*
From the above equations

*λ/λ*_{1}*=*2/3
Since the frequency is inversely proportional to the wave length we have

*λ/λ*

_{1}

*=*

*f*

_{1}/

*f*

Therefore

*f*_{1}/*f*= 2/3 from which*f*_{1 }= 2*f/*3
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